Download Presentation
## Warm Up

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Warm Up**Jessica is standing 75 feet from the base of a vertical tree. She is 5.5 feet tall and her eyes are 0.35 feet from the top of her head. It takes a 32º angle of elevation for Jessica to look at the top of the tree. Find the height of the tree in feet.**Answer**Height of Jessica’s eyes 5.5 – 0.35 = 5.15 x 32° 5.15 ft 75 ft cos32 = 75 x x = 88.43838025 Tree height 88.43838025 + 5.15 = 93.58838025 feet The height of the tree is 94 feet.**Bearing Word Problems**Bearing Word Problems are an application of trigonometry that is used in navigation. Caution: Angles in Bearing Problems are not measured from standard position.**Bearing**There are two methods for expressing bearing. When a single angle is given, such as 164˚, it is understood that the bearing is measured in a clockwise direction from due north.**Bearing**The second method for expressing bearing starts with a North-South line and uses an acute angle to show the direction, either East or West, from this line.**Radar stations A and B are on an east-west line, 8.6 km**apart. Station A detects a plane at C, on a bearing of 53˚. Stations B simultaneously detects the same plane, on a bearing of 323˚. Find the distance from B to C. Example 1a: SOLVING A PROBLEM INVOLVING BEARING (FIRST METHOD) A line drawn due north is perpendicular to an east-west line, so right angles are formed at A and B, and angles CAB and CBA can be found as shown in the figure. ∠C is a right angle because ∠CAB and ∠CBA are complementary.**Find distance a by using the cosine function for angle A.**Example 1a The distance from B to C is about 5.2 km.**Example 1b**SOLVING A PROBLEM INVOLVING BEARING (FIRST METHOD) Radar stations A and B are on an east-west line, 3.7 km apart. Station A detects a plane at C, on a bearing on 61˚. Station B simultaneously detects the same plane, on a bearing of 331˚. Find the distance from A to C.**Caution**A correctly labeled sketch is crucial when solving bearing applications. Some of the necessary information is often not directly stated in the problem and can be determined only from the sketch.**Example 2a:**SOLVING PROBLEM INVOLVING BEARING (SECOND METHOD) The bearing from A to C is S 52˚E. The bearing from A to B is N 84˚E. The bearing from B to C is S 38˚W. A plane flying at 250 mph takes 2.4 hours to go from A to B. Find the distance from A to C. To draw the sketch, first draw the two bearings from point A. Choose a point B on the bearing N 84˚ E from A, and draw the bearing to C, which is located at the intersection of the bearing lines from A and B.**Example 2a**SOLVING PROBLEM INVOLVING BEARING (SECOND METHOD) (cont.) The distance from A to B is about 430 miles.**Example 2b**The information in the example gives and The bearing from A to C is N 64˚W. The bearing from A to B is S 82˚W. The bearing from B to C is N 26˚E. A plane flying at 350 mph take 1.8 hours to go from A to B. Find the distance from B to C.**Example 2b**SOLVING PROBLEM INVOLVING BEARING (SECOND METHOD) The sum of the measures of angles EAB and FBA is 180º because they are interior angles on the same side of a transversal. 90° 34° 56° 98°**Example 2b**630 mi SOLVING PROBLEM INVOLVING BEARING (SECOND METHOD) It takes 1.8 hours at 350 mph to fly from A to B, so 90° 34° 56° 98° The distance from B to C is about 350 miles.